'how to get all data in correct way firebase-real-time-database JavaScript

I am using node.js and I am getting data from firebase real-time database. The problem is I am getting data something like this :

for data getting code! JS

import firebaseApp from '../config.js';
import { getDatabase, ref, onValue } from "firebase/database";


const userRef = ref(database, "Users");
onValue(userRef, (snapshot) => {

if (snapshot.exists) {
  const data = snapshot.val();
  console.log(data); // data printed to console
}

}, {
  onlyOnce: true
});

Console Output

{
  
 "random-user-id-1": {
    "name": "Jhon Doe",
    "profile": "profilelink",
    "email": "[email protected]"
 },
 
 "random-user-id-2": {
    "name": "Cr7",
    "profile": "profilelink",
    "email": "[email protected]"
 },

 // and more...

}

I want to display this data as an array of objects. Example of expected output

[

    {
    "name": "Jhon Doe",
    "profile": "profilelink",
    "email": "[email protected]"
    },
    
    {
    "name": "Cr7",
    "profile": "profilelink",
    "email": "[email protected]"
    }

    // and more........ ^_~

]
Any help will be always appreciated! and feel free to ask any doubts related to my question or problem!

thank you :)



Solution 1:[1]

seems like you need only the values from your dict, you can transform your data this way:

const lst = {
  
 "random-user-id-1": {
    "name": "Jhon Doe",
    "profile": "profilelink",
    "email": "[email protected]"
 },
 
 "random-user-id-2": {
    "name": "Cr7",
    "profile": "profilelink",
    "email": "[email protected]"
 },

}

const expectedFormatRes = Object.values(lst);


console.log(expectedFormatRes);

Solution 2:[2]

An alternative to Gil's answer would be to use Firebase's built-in. forEach operation:

if (snapshot.exists) {
  let values = [];
  snapshot.forEach((child) => {
    value.push(child.val());
  })
  console.log(values);
}

While longer, this has the advantage that it maintains the order in which the database returned the data, which becomes relevant when you specify an orderBy... clause on your query..

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 gil
Solution 2 Frank van Puffelen